Answer:
X₂ = 6.74 mm
Explanation:
As we know that:
ω[tex]_{n1}[/tex] = 900 (2π/60) rad/ sec = natural frequency
ω = 1750(2π/60) rad/ sec = maximumm frequency
Also,
r ₁ = ω / ω[tex]_{n1}[/tex] = [1750(2π/60)]/[900(2π/60)] = 1.94
Now we know that general formula for stiffness is
ω[tex]_{n1}[/tex] = [tex]\sqrt{k/m}[/tex] ---------- equation (1)
Similarly
ω[tex]_{n2}[/tex] = [tex]\sqrt{k/m}[/tex] ---------- equation (2)
put k = k/2 in equation 2, we get
[tex]\sqrt{2}[/tex]ω[tex]_{n2}[/tex] = [tex]\sqrt{k/m}[/tex]
Now, by equating equation 1 and 2, we get
ω[tex]_{n1}[/tex] = [tex]\sqrt{2}[/tex]ω[tex]_{n2}[/tex]
or
(1/[tex]\sqrt{2}[/tex])ω[tex]_{n1}[/tex] = ω[tex]_{n2}[/tex]
By putting ω[tex]_{n1}[/tex] = 900 (2π/60) we get
(1/[tex]\sqrt{2}[/tex])900 (2π/60) = ω[tex]_{n2}[/tex]
By solving it we get
ω[tex]_{n2}[/tex] = 636.4(2π/60)
Also we know that
r ₁ = ω / ω[tex]_{n2}[/tex]
By putting values in it, we get
r₂ = [1750(2π/60)] / [636.4 (2π/60)]
r₂ = 2.75
For τ = 0, we have rotating unbalanced equation as
X =[m[tex]_{u}[/tex]∈/m][r²/(1-r²)]
For X=8, r = r₁ ; 8 =[m[tex]_{u}[/tex]∈/m][r₁²/(1-r₁²)] ---------------equation (a)
For X=X₂, r = r₂ ; 8 =[m[tex]_{u}[/tex]∈/m][r₂²/(1-r₂²)] ---------------equation (b)
Dividing equation a by b , we get
X₂/8 = (r₂/r₁)²[(1-r₁²)/(1-r₂²)]
Now, put r₁ = 1.94 ; r₂ = 2.75
we get
X₂/8 = (2.75/1.94)²[(1-1.94²)/(1-2.75²)]
= 0.842
X₂ = 6.74 mm
The amplitude of one half thickness is calculated to be 7.74mm
